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Math Help - Verifying Identities

  1. #1
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    Exclamation Verifying Identities

    (1+cscx/secx) - cotx = cosx



    secx + tanx = (cosx/1-sinx)



    (sinx/1-cosx) = (1+cosx/sinx)
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  2. #2
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    Quote Originally Posted by nichole23 View Post
    (1+cscx/secx) - cotx = cosx
    Hi nichole23,

    Try posting these one at a time. I think you'd have a better chance of getting them answered. Just a suggestion.


    \frac{1+\csc x}{\sec x}-\cot x=\cos x


    \dfrac{1+\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}}-\dfrac{\cos x}{\sin x}=


    \dfrac{\dfrac{\sin x+1}{\sin x}}{\dfrac{1}{\cos x}}-\dfrac{\cos x}{\sin x}=


    \dfrac{\cos x(\sin x+1-1)}{\sin x}=


    \dfrac{\cos x(\sin x)}{\sin x}=


    \cos x
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  3. #3
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    Quote Originally Posted by nichole23 View Post

    secx + tanx = (cosx/1-sinx)
    Hi again nichole23,

    \sec x+\tan x=\frac{\cos x}{1-\sin x}

    \frac{1}{\cos x}+\frac{\sin x}{\cos x}=\frac{\cos x}{1-\sin x}

    \frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x}

    Multiply left side by \frac{1-\sin x}{1-\sin x}

    \frac{1-\sin^2 x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}

    \frac{\cos^2 x}{\cos x(1-\sin x)}=\frac{\cos x}{1-\sin x}

    \frac{\cos x}{1-\sin x}=\frac{\cos x}{1-\sin x}
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  4. #4
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    Quote Originally Posted by nichole23 View Post

    (sinx/1-cosx) = (1+cosx/sinx)
    Hello again,


    \frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}


    Multiply left side by \frac{1+\cos x}{1+\cos x}


    \frac{\sin x(1+\cos x)}{1-\cos^2 x}=\frac{1+\cos x}{\sin x}


    \frac{\sin x(1+\cos x)}{\sin^2 x}=\frac{1+\cos x}{\sin x}


    \frac{1+\cos x}{\sin x}=\frac{1+\cos x}{\sin x}
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  5. #5
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    Thank you so much!!!! I really appreciate it!

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